hdu【1712】ACboy needs your help
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ACboy needs your help
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5878 Accepted Submission(s): 3201
Problem Description
ACboy has N courses this term, and he plans to spend at most M days on study.Of course,the profit he will gain from different course depending on the days he spend on it.How to arrange the M days for the N courses to maximize the profit?
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers N and M, N is the number of courses, M is the days ACboy has.
Next follow a matrix A[i][j], (1<=i<=N<=100,1<=j<=M<=100).A[i][j] indicates if ACboy spend j days on ith course he will get profit of value A[i][j].
N = 0 and M = 0 ends the input.
Next follow a matrix A[i][j], (1<=i<=N<=100,1<=j<=M<=100).A[i][j] indicates if ACboy spend j days on ith course he will get profit of value A[i][j].
N = 0 and M = 0 ends the input.
Output
For each data set, your program should output a line which contains the number of the max profit ACboy will gain.
Sample Input
2 21 21 32 22 12 12 33 2 13 2 10 0
Sample Output
346
Source
HDU 2007-Spring Programming Contest
P06: 分组的背包问题
问题
有N件物品和一个容量为V的背包。第i件物品的费用是c[i],价值是w[i]。这些物品被划分为若干组,每组中的物品互相冲突,最多选一件。求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量,且价值总和最大。
算法
这个问题变成了每组物品有若干种策略:是选择本组的某一件,还是一件都不选。也就是说设f[k][v]表示前k组物品花费费用v能取得的最大权值,则有:
f[k][v]=max{f[k-1][v],f[k-1][v-c[i]]+w[i]|物品i属于组k}
使用一维数组的伪代码如下:
for 所有的组k
for v=V..0
for 所有的i属于组k
f[v]=max{f[v],f[v-c[i]]+w[i]}
注意这里的三层循环的顺序,甚至在本文的第一个beta版中我自己都写错了。“for v=V..0”这一层循环必须在“for 所有的i属于组k”之外。这样才能保证每一组内的物品最多只有一个会被添加到背包中。
#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int maxn = 105;int dp[maxn];int A[maxn][maxn];int main(){ int n,m; while(scanf("%d%d",&n,&m) != EOF) { if(!n && !m) break; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) scanf("%d",&A[i][j]); memset(dp,0,sizeof(dp)); for(int i = 1; i <= n; i++) for(int j = m; j >= 0; j--) for(int k = 1; k <= m; k++) if(j >= k) dp[j] = max(dp[j],dp[j-k] + A[i][k]); printf("%d\n",dp[m]); } return 0;}
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